A small mass `m` is moved slowly from the surface of the earth to a height `h` above the surface. The work done (by an external agent) in doing this is

Options

A. `mgh`, for all values of h
B. `mgh`, for `hlt ltR`
C. `1//2, mgR`, for `h=R`
D. `-1//2mgR`, for `h=R`

Explanation

To solve the problem of calculating the work done by an external agent in moving a small mass `m` from the surface of the Earth to a height `h`, we will follow these steps:

### Step 1: Understand the concept of gravitational potential energy
The gravitational potential energy (U) at a distance `r` from the center of the Earth is given by the formula:
\[ U = -\frac{G M m}{r} \]
where:
- \( G \) is the gravitational constant,
- \( M \) is the mass of the Earth,
- \( m \) is the mass of the object,
- \( r \) is the distance from the center of the Earth.

### Step 2: Calculate the potential energy at the surface of the Earth
At the surface of the Earth, the distance \( r \) is equal to the radius of the Earth \( R \):
\[ U_1 = -\frac{G M m}{R} \]

### Step 3: Calculate the potential energy at height \( h \)
At a height \( h \) above the surface, the distance from the center of the Earth becomes \( R + h \):
\[ U_2 = -\frac{G M m}{R + h} \]

### Step 4: Calculate the work done by the external agent
The work done (W) by the external agent in moving the mass from the surface to height \( h \) is the change in potential energy:
\[ W = U_2 - U_1 \]
Substituting the values from Steps 2 and 3:
\[ W = \left(-\frac{G M m}{R + h}\right) - \left(-\frac{G M m}{R}\right) \]
\[ W = \frac{G M m}{R} - \frac{G M m}{R + h} \]

### Step 5: Simplify the expression for work done
Factoring out \( G M m \):
\[ W = G M m \left(\frac{1}{R} - \frac{1}{R + h}\right) \]
To combine the fractions:
\[ W = G M m \left(\frac{(R + h) - R}{R(R + h)}\right) \]
\[ W = G M m \left(\frac{h}{R(R + h)}\right) \]

### Step 6: Consider the case when \( h \) is very small compared to \( R \)
If \( h \) is much smaller than \( R \) (i.e., \( h \ll R \)), we can approximate:
\[ W \approx \frac{G M m h}{R^2} \]
Since \( g = \frac{G M}{R^2} \), we can write:
\[ W \approx mgh \]

### Step 7: Consider the case when \( h \) is equal to \( R \)
If \( h = R \), substituting \( h \) into the work done expression:
\[ W = G M m \left(\frac{R}{R^2}\right) = \frac{G M m}{R} \]
This can be simplified further, but for practical purposes, we can conclude that the work done increases as \( h \) increases.

### Final Result
Thus, the work done by an external agent in moving the mass \( m \) to height \( h \) is given by:
\[ W = \frac{G M m h}{R(R + h)} \]
And for small heights, it approximates to \( W \approx mgh \).